Mathematics Grade 10 KZN March Test 2024 Paper & Memo

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE a a u nu “ " " un a u MARKS: 75 TIME: 1% hours This question paper consists of 5 pages.

i March 2024 Common Test Mathematics 2 ownloaded from Stanmorephysics.com INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. I This question paper consists of 3 questions. 2: Answer ALL the questions. 3: Number the answers correctly according to the numbering system used in this question paper. 4. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. a Answers only will NOT necessarily be awarded full marks. 6. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. Ti If necessary, round off answers correct to TWO decimal places, unless stated otherwise. 8. Diagrams are NOT necessarily drawn to scale. 9. Write neatly and legibly. Copyright Reserved Please Turn Over

Mathematics Downloaded from Stanmorephysics.com QUESTION 1 | 12 1.3 1.4 V3) Indicate whether each of the following numbers is rational or irrational: LLU) yas 1.1.2--2,5 113 3g 1.1.4 2,3333333.... Write down TWO rational numbers between V2 and vio (Show your working). Write down 0,11111111.... as a fraction. Factorise the following expressions fully; Simplify the following expressions fully; 1.5.1 (2x-1)(x?-2x+1) 152 (3x+5)° 133 g~9** Qt _ 9s 154 3 2 21 a-4 a+3 a’-a-12 Copyright Reserved March 2024 Common Test (1) () qd) (1) (2) (2) (2) G3) (3) (2) (2) (4) (5) [29] Please Turn Over

Mathematics March 2024 Common Test Downloaded from Stanmorepitysics.com QUESTION 2 2.1 Solve for x: 211 (x+3)(x-1)=0 (2) 2255? —5x4.2=0 (3) BT3—~(o)" =64 (3) 214 97 agp (4) 2.2 Solve the inequality —2<4+2x<6 and represent the solution in the interval notation. (4) id y simultaneously if: (5) e retice between two numbers is 5. Six times the smaller number is equal to four times e yo number. Find the numbers. (4) 235 r If p=1+3* and g=1+3™ , prove that ge neal (3) [28] Copyright Reserved Please Turn Over

Mathematics 5 March 2024 Common Test Downloaded from Stanmorepfif¥sics.com QUESTION 3 3.1 A rightangles triangle with sides P, q and s and the angle @, as shown below. i . q (c) 3.1.1 Write down the values of p, g and s: (a) tano (1) (b) -sing () (c) sec? @ (2) 3.1.2 If itis given that p=12 and @=35° , calculate the numerical value of g. (2) 3.2. If A=20°and B=55° use your calculator to evaluate the following(correct to TWO decimal places). (2) (2) (3) 3.3. | Without the use of the calculator, showing all your working, determine the value of: + 15° tan2 EO sin 45 60' +sin30° cos (5) [18] TOTAL [75] Copyright Reserved

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE x= MARKS: 75 This marking guideline consists of 6 pages. Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics. comommon Test March 2024 GRADE 10 Marking Guideline QUESTION 1 1.1.1 | Irrational v answer () 1.1.2 | Rational ¥ answer (1) 1.1.3 | Rational ¥ answer (dd) 1.1.4 | Rational v answer () 12 | J4 and J9 Vv V4 and J9 =2 and3 ¥ both answers (2) 1.3 Let 0,11111....=k 1,11111.....=10k ¥1,11111.....=10k 9k =1 1 k=— 1 Y kei 2 9 (2) 14.1 =(x=2)(x? +2x+4) v (x-2) v (x? +2x+4) Q) P| er hy 2 x +1) ¥ Veach term (2 v answer (3) cit =(x-2)(x+2)(x? +4) Vv (x=2)(x+2)(x? +4) (3) 1.5.1 | =2x3~4x? 42x—x?42x-1 ¥ simplification = 3 ll 2 —_ = 2x" —5x° +4x-1 answer | Answer only: full marks (2) 1.5.2 =(3x+5)(3x+ 5) ¥ expansion ee 2 eae ees ¥ 9x? +15x+15x-+25 = 9x? +30x+25 ¥ answer (3) 1.5.3 2x 2* 27? Y factors ~ BEDLDE x wf = 2 (1-2 ) ¥ common factor 2* (2-1) 1 = IZ vi 3 4 3 bn (4) Copyright Reserved Please turn over

Mathemfewnloaded from Stanmorephysics.comommon Test March 2024 GRADE 10 Marking Guideline 1.5.4 _Ie3 = 2 21 a-4 a+3 (a-4)(a+3) _3(a+3)+2(a—4)-21 Y factorisation of a quadratic trinomial Y finding the LCD and the (a-4)(a+3) simplification of the numerator _ 3a+9+2a-8-21 (a—4)(a+3) 5a —20 ¥ simplification ~ (a4) (a+3 (a~4)(a+3) 6 = 5(a-4) Y factorising the numerator (a-4)(a+3) =o v answer a+3 [29] QUESTION 2 2.1.1 | x=-3 or x=1 ¥ V each answer (2) 23102 (2x-1)(x-2)=0 Y factors x= 2 8 x=2 ¥ ¥ each answer (3) 2:3 | 199x198 v 28 3x-1=6 y , ; —_ equating exponents wel 3 v answer (3) 2.1.4 ( 3\*-2 42x41 3\*-2 4 \2x41 3’) =(3") v(3) =(3‘) Es = aaa Ve a ; 3x-6=8x+4 simplification x=-2 ¥ equating exponents (4) ¥ answer 2.2 | -6<2x<2 ¥ transposing —3<xsl1 ef xe(-3;1] —3<x<1 ¥ ¥each end (4) 2.3 ¥ making x the subject of the formula Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics. comommon Test March 2024 GRADE 10 Marking Guideline Substitute eq. (3) into eq. (2) 2(5-3y)-6y+2=0 Y substitution 10-6y~6y+2=0 ¥ simplification y -value WF x=S—3(1) v x-value x=2 (2;1) () OR X43Y—5 =Oneeees > (1) 2X —BY +2 =Ovvrrrrrersesrees > (2) ¥ multiplying eq(1) by —2 Multiply eq. (1) by -2 -2x-6y +10=0......... > (3) 2X—6Y +2 =O... (2) V eliminating x Add eq. (3) and eq. (4) ¥ simplification 12y-12=0 y= v y-value =5-3(1 x ©) v x-value x=2 231 (2:1) (3) 2.4 | Let the smaller number be: x and the bigger number be: y VX See eceecee cee >(1) 6x=4y oe. (2) From eq(1) y= X45. serteeeed (3) ¥ y-subject of formula Subst. (3) into (2) 6x = 4(x+5) ¥ substitution 6x = 4x +20 V , x=10 X-value y=ls vy-value OR (4) From eq(1) X= V—S.eeeceeeee > (3) y¥ x-subject of the formula Subst. (3) into (2) V substituti 6(y-5)=4y substitution 6y-30=4y y=l5 vy-value x=10 v¥x-value (4) OR Copyright Reserved Please turn over

Mathemaawnloaded from Stanmorephysics.comommon Test March 2024 GRADE 10 Marking Guideline From eq(2) Th i —~ >(3) ¥ y-subject of formula Subst. eq(3) into eq(1) 3 =x=xs5 ¥ substitution Tyas 2 x=10 ¥x-vali =15 x-value y vy-value (4) 2.5 x ¥ substitution RHS = 1+3 1+3*-1 _1+3* 3* x 1 3 1,3 ary 3 ¥3*41 =3*+4+1 OR LHS =1 1 a3 V finding the L.C.D _ 341 3 rus = + 3* ¥ substitution 1+3*-1 _1+3° 3* Vv answer LHS = RHS 8) [28] QUESTION 3 3.1.1(a) s tan 6 = q v answer (1) 3.1.1(b) . s ~sind = “D v answer (1) 3.1.1(¢) p 2 ¥ substitution = [2] ¥ answer (2) q | Answer only: full marks _ | Pe ¢ Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.comommon Test March 2024 GRADE 10 Marking Guideline Copyright Reserved 3.1.2 Wr ESAT i> V trig. ratio = 9)83 ! v answer (2) 3.2.1 | =sin (20° ip 55‘) ¥ substitution v answer (2) = 0,97 | Answer only: full marks | 3222 | = (tan 55°) ¥ substitution v answer (2) = 2,04 3.2.3 | =2cosec20°+sin5(55°) ¥ substitution =— +sin5(55°) sin 20° =4,85 (3) 3.2.4 v2 3 2 Ns V2 2 2 ‘ 1 1 =3= v= 5 5 5 (5) ¥ answer [18] TOTAL: 75